Question

Both $\mathbf{\text{Ni}}{{\mathbf{\left(}{\mathbf{\text{NH}}}_{\mathbf{\text{3}}}\mathbf{\right)}}_{\mathbf{\text{4}}}}^{\mathbf{\text{2+}}}$ and${{\mathbf{\text{Ni(CN)}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{2-}}}$ have four ligands. The first is paramagnetic and the second is diamagnetic. Are the complex ions tetrahedral or square planar? Explain.

Short answer

Tetrahedral is$\text{Ni}{{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}}^{\text{2+}}$ and $\text{Ni}{{\left(\text{CN}\right)}_{\text{4}}}^{\text{2-}}$is square planar.

Step-by-step solution

A concept:

The crystal field splitting is the difference in energy between the d orbitals of the ligands. The division number of the crystal field is denoted by the capital Greek letter $\mathbf{∆}$. Crystal field splitting explains the difference in color between two similar metal-ligand complexes.

Explain the complex ions tetrahedral or square planar or not:

Here, ${\text{Ni}}^{2+}$is a ${\text{d}}^{\text{8}}$electronic species. ${\text{NH}}_{\text{3}}$is a weak field ligand whereas ${\text{CN}}^{-}$is a strong field ligand.

The strong field ligand changes the splitting pattern for${\text{d}}^{\text{8}}$ electronic species hence it forms square planar complex.